How can you have 0 as the difference of two squares? 5^2-5^2?
The one that doesn't follow the pattern of a^2 - b^2.
2m2 - 8 = 2(m2 - 4) = 2(m + 2)(m - 2)
It depends what you mean.a2-b2 is the difference of two squares so it equals (a-b)(a+b) (a-b)2=a2-2ab+b2if your question is a difference of squares and it looks like (a-b)^2, then your answer is (a+b)(a-b).
If you mean: (y-2)(y+2) then it is y^2 -4 which is the difference of two squares
How can you have 0 as the difference of two squares? 5^2-5^2?
difference of squares if something of the form a^2-b^2. So for example x^2-y^2 since both are squares. The value in looking at these is that we can factor a^2-b^2 in (a+b)(a-b)
a^2 - b^2 = (a + b)(a + b).
a^2 - b^2 = (a + b)(a - b)
x^2 - 64.
The difference of two squares: 4 -9 = (2-3)(2+3)
The one that doesn't follow the pattern of a^2 - b^2.
2m2 - 8 = 2(m2 - 4) = 2(m + 2)(m - 2)
It depends what you mean.a2-b2 is the difference of two squares so it equals (a-b)(a+b) (a-b)2=a2-2ab+b2if your question is a difference of squares and it looks like (a-b)^2, then your answer is (a+b)(a-b).
To factor a^4 - b^4 completely, you can use the formula for the difference of squares, which states that a^2 - b^2 = (a + b)(a - b). In this case, a^4 - b^4 is a difference of squares twice: (a^2)^2 - (b^2)^2. So, you can factor it as (a^2 + b^2)(a^2 - b^2). Then, factor a^2 - b^2 further using the difference of squares formula to get (a^2 + b^2)(a + b)(a - b), which is the complete factorization of a^4 - b^4.
If you mean: (y-2)(y+2) then it is y^2 -4 which is the difference of two squares
GFYM